Complex Number: What's the set?

[latentcorpse]

What's the set $$\{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}$$?

I've set z=a+ib and found $$a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)$$

I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find $$\{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}$$

How do we draw this?

Thanks.

 

[Mark44]

What's the set $$\{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}$$?

I've set z=a+ib and found $$a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)$$

$a^2 - 2a + b^2 \geq 0$
Complete the square in the a terms, and see what you get.

I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find $$\{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}$$

How do we draw this?

Thanks.

 

[Ray Vickson]

What's the set $$\{ z \in \mathbb{C}| |z|^2 \geq z+ \bar{z} \}$$?

I've set z=a+ib and found $$a^2 + b^2 \geq 2a \Rightarrow b^2 \geq a(2-a)$$

I'm not sure how to interpret this geometrically ie what it looks like?

I suppose it is the set of vectors whose length is bigger than twice their real part. I guess if I take the square root then I find $$\{ b \geq \pm \sqrt{a} \} \cap \{ b \geq \pm \sqrt{2-a} \}$$

How do we draw this?

Thanks.

Interpret $a^2 -2a + b^2 \geq 0$.

 

[latentcorpse]

so
$$(a-1)^2+(b-0)^2-1 \geq 0$$

$$(a-1)^2+(b-0)^2 \geq 1$$

so circle radius 1 centre (1,0)?

 

[Mark44]

Yes.

 

[latentcorpse]

Yes.

what was wrong with saying the length was bigger than twice the real part?

 

[vela]

Nothing, but which description is easier to visualize?

 

[HallsofIvy]

so
$$(a-1)^2+(b-0)^2-1 \geq 0$$

$$(a-1)^2+(b-0)^2 \geq 1$$

so circle radius 1 centre (1,0)?

$$(a- 1)^2+ b^2= 1$$
is a circle of radius 1 with center at (1, 0).

$$(a- 1)^2+ b^2\ge 1$$
is the set of points outside that circle.

 

[Mark44]

$$(a- 1)^2+ b^2= 1$$
is a circle of radius 1 with center at (1, 0).

$$(a- 1)^2+ b^2\ge 1$$
is the set of points outside that circle.

The inequality represents the set of points on the circle or outside it.